Problem: Let $f(x)=2x+1$. Find the sum of all $x$ that satisfy the equation $f^{-1}(x)=f(x^{-1})$.
Solution: In order to find $f^{-1}$ we substitute $f^{-1}(x)$ into our expression for $f$.  This gives  \[f(f^{-1}(x))=2f^{-1}(x)+1.\]Since $f(f^{-1}(x))=x$, this equation is equivalent to  \[x=2f^{-1}(x)+1,\]which simplifies to  \[f^{-1}(x)=\frac{x-1}2.\]If we assume $x$ solves $f^{-1}(x)=f(x^{-1})$, then we get  \[\frac{x-1}2=\frac 2x+1=\frac{2+x}x.\]Cross-multiplying gives  \[x^2-x=4+2x.\]Then $x^2 - 3x - 4 = 0$.  Factoring gives $(x-4)(x+1)=0$, from which we find $x=4$ or $x=-1$.  The sum of the solutions is $4+(-1) = \boxed{3}$.

Alternatively, since Vieta's formula tells us that the sum of the roots of a quadratic $ax^2+bx+c$ is $-\frac{b}{a}$, the sum of the roots of $x^2-3x-4$ is $-\frac{-3}{1}=\boxed{3}$.